The water tank in the diagram below is in the shape of an inverted right circular cone. The radius of its base is 16 feet, and its height is 96 feet. The water in the tank is 50% of the tank's capacity. Find the height of the water in the tank.
Volume of water held when the tank is full
(1/3) pi * radius ^2 * height = (1/3) pi * (16)^2 (96) = 8192 pi ft^3
When the tank is half- full.....the volume = 8192 pi / 2 = 4096 pi ft^3
Using silimlae triangles we have that
r / h = 16 / 96
r / h = 1/6
r = h/6
So.......we have that
4096 pi = (1/3) pi r^2 h subbing for r we have that
4096 pi = (1/3) pi ( h/6)^2 * h simplify
4096 = (1/3) h^3 / 36
4096 = h^3 / 108
108 * 4096 = h^3
442368 = h^3 take the cube root of both sides
h ≈ 76.2 ft